Question

A bomb at rest explodes into three parts of same mass. The momentum of two parts is \( -3 \, P_i \) and \( 2 \, P_j \) respectively. The magnitude of the momentum of third part is

• A. \( \sqrt{11} \, P \)
• B. \( \sqrt{13} \, P \)
• C. \( \sqrt{15} \, P \)
• D. \( \sqrt{7} \, P \)

Hint:

In explosions, the total momentum of the system remains zero, and the magnitudes of the momenta of the parts can be related using the Pythagorean theorem.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
Since the bomb is initially at rest, the total momentum of the system is zero. After the explosion, the vector sum of the momenta of the three parts must still be zero. \[ \vec{P_1} + \vec{P_2} + \vec{P_3} = 0 \] Step 2: Apply the given momentum values.
The momentum of the first two parts is given as \( \vec{P_1} = -3P_i \) and \( \vec{P_2} = 2P_j \). Let the momentum of the third part be \( \vec{P_3} = P_3 \). Step 3: Apply the Pythagorean theorem.
Since the total momentum is zero, we can use the Pythagorean theorem to find the magnitude of the third momentum \( P_3 \). We have: \[ P_3 = \sqrt{P_1^2 + P_2^2} \] Substituting the given values: \[ P_3 = \sqrt{(-3P)^2 + (2P)^2} = \sqrt{9P^2 + 4P^2} = \sqrt{13P^2} = \sqrt{13} P \] Step 4: Conclusion.
Thus, the magnitude of the momentum of the third part is \( \sqrt{13} P \), which corresponds to option (B).